Theorem nss 3986

Description: Negation of subclass relationship. Exercise 13 of [TakeutiZaring] p. 18. (Contributed by NM, 25-Feb-1996.) (Proof shortened by Andrew Salmon, 21-Jun-2011.)

Assertion

Ref	Expression
nss	⊢ (¬ 𝐴 ⊆ 𝐵 ↔ ∃𝑥(𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵))

Distinct variable groups:   𝑥,𝐴   𝑥,𝐵

Proof of Theorem nss

Step	Hyp	Ref	Expression
1		exanali 1861	. . 3 ⊢ (∃𝑥(𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵) ↔ ¬ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
2		df-ss 3906	. . 3 ⊢ (𝐴 ⊆ 𝐵 ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
3	1, 2	xchbinxr 335	. 2 ⊢ (∃𝑥(𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵) ↔ ¬ 𝐴 ⊆ 𝐵)
4	3	bicomi 224	1 ⊢ (¬ 𝐴 ⊆ 𝐵 ↔ ∃𝑥(𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 206   ∧ wa 395  ∀wal 1540  ∃wex 1781   ∈ wcel 2114   ⊆ wss 3889

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1782  df-ss 3906

This theorem is referenced by:  grur1  10743  psslinpr  10954  reclem2pr  10971  mreexexlem2d  17611  prmcyg  19869  filconn  23848  alexsubALTlem4  24015  wilthlem2  27032  shne0i  31519  onvf1odlem2  35286  erdszelem10  35382  fundmpss  35949  ntrneineine1lem  44511  nssrex  45516  nssd  45535  nsssmfmbf  47207